Chapter: Mechanical - Dynamics of Machines - Force Analysis

Inertia Forces in a Reciprocating Engine, Considering the Weight of connecting Rod

In a reciprocating engine, let OC be the crank and PC, the connecting rod whose centre of gravity lies at G. The inertia forces in a reciprocating engine may be obtained graphically as discussed below:

INERTIA FORCES IN A RECIPROCATING
ENGINE, CONSIDERING THE WEIGHT OF CONNECTING ROD

In a
reciprocating engine, let OC be the
crank and PC, the connecting rod
whose centre of gravity lies at G.
The inertia forces in a reciprocating engine may be obtained graphically as
discussed below:

1. First
of all, draw the acceleration diagramOCQNby Klien’s construction. Weknow
that the acceleration of the piston P
with respect to O,

acting in the direction from N to O.
Therefore, the inertia force FI
of the reciprocating parts will act in the opposite direction as shown in Fig.
15.22.

2. Replace
the connecting rod by dynamically equivalent system of two masses as discussed
in Art. 15.12. Let one of the masses be arbitrarily placed at P. To obtain the position of the other
mass, draw GZ perpendicular to CP such that GZ = k, the radius of
gyration of the connecting rod. Join PZ
and from Z draw perpendicular to DZ which intersects CP at D. Now, D is the position of the second mass.

Note: The position of the second mass may
also be obtained from the equation,

GP
×
GD = k2

3. Locate the pointsGandDonNCwhich is the
acceleration image of theconnecting
rod. This is done by drawing parallel lines from G and D to the line of
stroke PO. Let these parallel lines
intersect NC at g and d respectively.
Join gO and dO. Therefore,acceleration
of G with respect to O, in the direction from g to O,

aGO = aG = ω2 × gO

and acceleration of D with respect to O, in the direction from d
to O,

aDO =
aD = ω2 × dO

From D, draw DE parallel to dO which intersects the line of stroke PO at E. Since the accelerating forces on the masses at P and D intersect at E, therefore
their resultant must also pass through E.
But their resultant is equal to the accelerang force on the rod, so that the
line of action of the accelerating force on the rod, is given by a line drawn
through E and parallel to gO, in the direc- tion from g to O.
The inertia force of the connecting rod FCtherefore acts through E and in the opposite direction as
shown in Fig. 15.22. Theinertia
force of the connecting rod is given by

A little consideration will show that the forces acting on
the connecting rod are :

(a) Inertia force of the reciprocating
parts (FI ) acting along
the line of stroke PO,

(b) The side thrust between the
crosshead and the guide bars (FN)
acting at P and

Now, produce the lines of action of FR and FN to intersect at a point I, known as instantaneous centre. From I draw I X and I Y , perpendicular to the lines of
action of FC and WC. Taking moments about I, we have

FT× IC = FI× IP + FC× I X + W C× I Y ...(ii)

The value of FT may be obtained from this equation and from the force
polygon as

shown in Fig. 15.22, the forces FN and FR may be calculated. We know that, torque exerted on
the crankshaft to overcome the inertia of the moving parts = FT × OC

1 Analytical Method for Inertia Torque

The effect of the inertia of the
connecting rod on the crankshaft torque may be obtained as discussed in the
following steps:

Fig. 15.23. Analytical method for inertia
torque.

1.The mass of the connecting rod (mC) is divided into two masses. One of the mass is
placed at the crosshead pin P and the
other at the crankpin C as shown in
Fig. 15.23, so that the centre of gravity of these two masses coincides with
the centre of gravity of the rod

G.

2.Since the inertia force due to the
mass at C acts radially outwards
along the crank OC, therefore the
mass at C has no effect on the
crankshaft torque.

3.The inertia force of the mass at P may be obtained as

follows: Let
connecting rod, mC = Mass of the

l =
Length of the connecting rod,

l1 =
Length of the centre of gravity of the connecting rod from P.

4. In
deriving the equation (ii) of the torque exerted on the
crankshaft, it isassumed that one
of the two masses is placed at C and
the other at P. This assumption does
not satisfy the condition for kinetically equivalent system of a rigid bar.
Hence to compensate for it, a correcting torque is neces- sary whose value is
given by

The correcting torque T' may be applied to the system by two
equal and opposite forces FYacting
through P and C. Therefore,

Q.The
crank and connecting rod lengths of an engine are 125 mm and 500 mmrespectively. The mass of the connecting rod is 60 kg and
its centre of gravity is 275 mm from the crosshead pin centre, the radius of
gyration about centre of gravity being 150 mm.

If the engine speed is 600 r.p.m. for a crank position of
45° from the inner dead centre, determine, using Klien’s or any other
construction 1. the acceleration of
the piston; 2. the magni- tude,
position and direction of inertia force due to the mass of the connecting rod.

First of all, draw the configuration
diagram OCP, as shown in Fig. 15.24,
to some suitable scale, such that

OC
=
r = 125 mm ; PC = l = 500 mm ; and= 45°.

Now, draw the Klien’s acceleration
diagram OCQN, as shown in Fig. 15.24,
in the same manner as already discussed. By measurement,

NO
= 90 mm = 0.09 m

 Acceleration of the piston,

aP=ω2× NO = (62.84)2
× 0.09 = 355.4 m/s Ans.

2. The magnitude,
position and direction of inertia force due to the mass of theconnecting rod

The magnitude, postition and
direction of the inertia force may be obtained as follows:

(i)
Replace the connecting rod by
dynamical equivalent system of two masses,assuming that one of the masses is placed at P and the other mass at D.
The position of the point D is
obtained as discussed in Art. 15.12.

(ii)
Locate the pointsGandDonNCwhich is the
acceleration image of theconnecting
rod. Let these points are g and d on NC.
Join gO and dO. By measurement,

gO
= 103 mm = 0.103 m

 Acceleration of G, aG=ω2× gO, acting in
the direction from g to O.

(iii)
From pointD, drawDEparallel todO. NowEis the point through which theinertia force of the connecting rod
passes. The magnitude of the inertia force of the connecting rod is given by

FC= mC×ω2× gO = 60 ×
(62.84)2× 0.103 = 24 400
N = 24.4kN Ans. (iv) From point E, draw a line parallel to gO, which shows the position of the

Q. The connecting rod of an internal
combustion engine is 225 mm long and has amass 1.6 kg. The mass of the piston
and gudgeon pin is 2.4 kg and the stroke is 150 mm. The cylinder bore is 112.5
mm. The centre of gravity of the connection rod is 150 mm from the small end.
Its radius of gyration about the centre of gravity for oscillations in the
plane of swing of the connect- ing rod is 87.5 mm. Determine the magnitude and
direction of the resultant force on the crank pin when the crank is at 40° and
the piston is moving away from
inner dead centre under an effective gas presure of 1.8 MN/m2. The
engine speed is 1200 r.p.m.